eSolutions Manual - Powered by Cognero Page 1 · Find the geometric mean between each pair of numbers. 5 and 20 62/87,21 By the definition, the geometric mean x of any two numbers

Find the geometric mean between each pair of numbers.

1.5 and 20

SOLUTION:By the definition, the geometric mean x of any two

numbers a and b is given by

Therefore, the geometric mean of 5 and 20 is

ANSWER:10

2.36 and 4




ANSWER:12

3.40 and 15




ANSWER:

or 24.5

4.Write a similarity statement identifying the three similar triangles in the figure.

SOLUTION:

If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to

the original triangle and to each other. isthealtitude to the hypotenuse of the right triangle CED.

Therefore,

ANSWER:

Find x, y, and z.

5.

SOLUTION:

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of this altitude is the geometric mean between thelengthsofthesetwosegments. Solve for x.

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of a leg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuseadjacenttothatleg. Solve for y .

Solve for z .

ANSWER:

x = 6; ; .

6.

SOLUTION:



By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of a leg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuseadjacenttothatleg. Solveforz.

ANSWER:

x = 32; ;

7.CCSS MODELING Corey is visiting the Jefferson Memorial with his family. He wants to estimate the height of the statue of Thomas Jefferson. Corey stands so that his line of vision to the top and base of the statue form a right angle as shown in the diagram. About how tall is the statue?

SOLUTION:We have the diagram as shown.

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of this altitude is the geometric mean between the lengths of these two segments.

So, the total height of the statue is about 159 + 68 or 227 inches, which is equivalent to 18 ft 11 in.

ANSWER:18 ft 11 in.


8.81 and 4




ANSWER:18

9.25 and 16




ANSWER:20

10.20 and 25




ANSWER:

11.36 and 24




ANSWER:

12.12 and 2.4



Therefore, the geometric mean of 12 and 2.4 is

ANSWER:

13.18 and 1.5




ANSWER:

Write a similarity statement identifying the three similar triangles in the figure.

14.

SOLUTION:If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.

isthealtitudetothehypotenuseoftherighttriangle MNO. Therefore,

ANSWER:

15.


isthealtitudetothehypotenuseoftherighttriangle XYW. Therefore,

ANSWER:

16.


isthealtitudetothehypotenuseoftheright

triangle QRS. Therefore,

ANSWER:

17.



triangle HGF. Therefore,

ANSWER:

18.

SOLUTION:



By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of a leg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuseadjacenttothatleg. Solve for z .

ANSWER:

x = 6; ;

19.

SOLUTION:

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of this altitude is the geometric mean between thelengthsofthesetwosegments. Solve for y .

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of a leg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So, Solve for x.

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of a leg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuseadjacenttothatleg. Solve for z.

ANSWER:

; ;

20.

SOLUTION:

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is the geometric mean between thelengthsofthesetwosegments. Solve for y .

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of aleg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuseadjacenttothatleg. Solve for x.

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of aleg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuseadjacenttothatleg. Solve for z.

ANSWER:

; ;

21.

SOLUTION:

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of aleg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuseadjacenttothatleg. Solve for y .

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of aleg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuseadjacenttothatleg. Use the value of y to solve the second proportion.

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is the geometric mean between the lengths of these two segments. So, Solve for x.

ANSWER:

x 4.7y 1.8z 13.1

22.

SOLUTION:

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is the geometric mean between thelengthsofthesetwosegments. Solve for x.

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of aleg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuse adjacent to that leg. Solve for y .

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of aleg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuse adjacent to that leg. Solve for z .

ANSWER:

x = 40; ;

23.

SOLUTION:


By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of aleg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuse adjacent to that leg. Use the value of z to solve the second proportion for x.


ANSWER:

; ;z = 32

24.CCSS MODELING Evelina is hanging silver stars from the gym ceiling using string for the homecomingdance. She wants the ends of the strings where the stars will be attached to be 7 feet from the floor. Usethe diagram to determine how long she should make the strings.

SOLUTION:Let x represent the length of the string. Since the starwill be 7 feet from the floor, x + 7 is the total length of string to floor. Since we are given 5 feet from the floor in the diagram. The distance to the 5 ft point will be x+2.

Use the Geometric Mean (Altitude) Theorem to find x.

So she should make the strings of length 18 feet.

ANSWER:18 ft

25.CCSS MODELING Makayla is using a book to sight the top of a waterfall. Her eye level is 5 feet from the ground and she is a horizontal distance of 28feet from the waterfall. Find the height of the waterfall to the nearest tenth of a foot.


By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of this altitude is the geometric mean between thelengthsofthesetwosegments.

So, the total height of the waterfall is 156.8 + 5 = 161.8 ft.

ANSWER:161.8 ft


26. and 60



Therefore, the geometric mean of and60is

ANSWER:

or 3.5

27. and



Therefore, the geometric mean of is

ANSWER:

or 0.8

28. and




ANSWER:

or 2.2

Find x, y, and z.

29.

SOLUTION:


Use the value of x to solve the for z .


ANSWER:

; ; z = 3

30.

SOLUTION:

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is the geometric mean between thelengthsofthesetwosegments. Solve for z .



ANSWER:

; ;

31.ALGEBRA The geometric mean of a number and four times the number is 22. What is the number?

SOLUTION:Let x be the first number. Then the other number willbe 4x. By the definition, the geometric mean x of any

two numbers a and b is given by So,

Therefore, the number is 11.

ANSWER:11

Use similar triangles to find the value of x.32.Refer to the figure on page 543.

SOLUTION:

By the Geometric Mean (Leg) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of aleg of this triangle is the geometric mean between thelength of the hypotenuse and the segment of the hypotenuse adjacent to that leg. Let y be the shorter segment of the hypotenuse of the bigger right triangle.

So, the shorter segment of the right triangle is about 14.2-4.1=10.1 ft. By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is the geometric mean between thelengthsofthesetwosegments.

ANSWER:6.4 ft

33.Refer to the figure on page 543.

SOLUTION:


So, the longer segment of the right triangle is about 3.98 ft. By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is the geometric mean between thelengthsofthesetwosegments.

The length of the segment is about 3.5 ft.

ANSWER:3.5 ft


SOLUTION:



Then x isabout13.5ft.

ANSWER:13.75 ft

ALGEBRA Find the value(s) of the variable.

35.

SOLUTION:

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is the geometric mean between thelengthsofthesetwosegments.

ANSWER:5

36.

SOLUTION:


Use the quadratic formula to find the roots of the quadratic equation.

If w = 16, the length of the altitude will be 16 + 4 =12 which is not possible, as a length cannot be negative. Therefore, w = 8.

ANSWER:8

37.

SOLUTION:



Since m is a length, it cannot be negative. Therefore, m = 4.

ANSWER:4

38.CONSTRUCTION A room-in-attic truss is a truss design that provides support while leaving area that can be enclosed as living space. In the diagram,

BCA and EGB are right angles, is

isosceles, is an altitude of , and is an

altitude of . If DB = 5 feet, CD = 6 feet 4 inches, BF = 10 feet 10 inches, and EG = 4 feet 6 inches, what is AE?

SOLUTION:

First find

Giventhat is isosceles, then bisectsSince is 10 ft 10 in.or 130 in., then BG =

GF = 65 in.

isarighttriangleandbythePythagoreanTheorem,

Next find AD. Let AD = x.

By the Geometric Mean (Altitude) Theorem the altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments and the length of this altitude is the geometric mean between the lengths of these two segments.

So, the total length AE is about (96.3 + 60 + 84.5) in. = 240.8 in. or 20.07 ft.

ANSWER:about 20.07 ft

CCSS ARGUMENTS Write a proof for each theorem.

39.Theorem 8.1

SOLUTION:

You need to walk through the proof step by step. Look over what you are given and what you need to prove. Here, you are given a right angle and an altitude of a triangle. Use the properties that you have learned about congruent segments, altitudes, right triangles, and equivalent expressions in algebra to walk through the proof.

Given: PQR is a right angle. isanaltitudeof

.

Prove:

Proof:

Statements (Reasons)

1. PQR is a right angle. isanaltitude

of . (Given)

2. (Definition of altitude)

3. 1 and 2 are right angles. (Definition of

perpendicularlines)

4. ; (Allrightanglesare

congruent.)

5. ; (Congruence of angles is

reflexive.)

6. ; AA (Similarity

Statements 4 and 5)

7. (Similarityoftrianglesis

transitive.)

ANSWER:

Given: PQR is a right angle. isanaltitudeof.

Prove:

Proof: Statements (Reasons)

1. PQR is a right angle. isanaltitudeof . (Given)


3. 1 and 2 are right s. (Definition of lines) 4. ; (Allright s are .)

5. ; (Congruence of angles is reflexive.)

6. ; AA (Similarity Statements 4 and 5)

7. (Similarity of triangles is transitive.)

40.Theorem 8.2

SOLUTION:

You need to walk through the proof step by step. Look over what you are given and what you need to prove. Here, you are given a right triangle and an altitude. Use the properties that you have learned about right triangles, altitudes, congruent segment,s and equivalent expressions in algebra to walk throughthe proof.

Given: isarighttriangle. isanaltitudeof.

Prove:

Proof: It is given that is a right triangle

and is an altitude of . ADC is a right

angle by the definition of a right triangle. Therefore,

, because if the altitude is drawn

from the vertex of the right angle to the hypotenuse

of a right triangle, then the two triangles formed are

similar to the given triangle and to each other. So

bydefinitionofsimilartriangles.

ANSWER:


Prove:


and is an altitude of . ADC is a right angle by the definition of a right triangle. Therefore,

, because if the altitude is drawn from the vertex of the right angle to the hypotenuse of a right triangle, then the two triangles formed are similar to the given triangle and to each other. So


41.Theorem 8.3

SOLUTION:You need to walk through the proof step by step. Look over what you are given and what you need to prove. Here, you are given a right triangle and an altitude. Use the properties that you have learned about congruent segments, right triangles, altitudes, and equivalent expressions in algebra to walk throughthe proof.

Given: ADC is a right angle. isanaltitudeof.

Prove: ;

Proof:


1. ADC is a right angle. is an altitude of

(Given)

2. is a right triangle. (Definition of right

triangle)

3. ; (If the altitude is

drawnfromthevertexoftherightangletothe

hypotenuse of a right triangle, then the 2 triangles

formedaresimilartothegiventriangleandtoeach

other.)

4. ; (Definition of similar

triangles)

ANSWER:


Prove: ;

Proof:



(Given)


triangle)


drawn from the vertex of the rt. tothe

hypotenuse of a rt. , then the 2 s formed are

similar to the given andtoeachother.)


triangles)

42.TRUCKS In photography, the angle formed by the top of the subject, the camera, and the bottom of the subject is called the viewing angle, as shown in the

diagram. Natalie is taking a picture of Bigfoot #5, which is 15 feet 6 inches tall. She sets her camera ona tripod that is 5 feet above ground level. The vertical

viewing angle of her camera is set for .


1.5 and 20




ANSWER:10

2.36 and 4




ANSWER:12

3.40 and 15




ANSWER:

or 24.5


SOLUTION:



Therefore,

ANSWER:

Find x, y, and z.

5.

SOLUTION:



Solve for z .

ANSWER:

x = 6; ; .

6.

SOLUTION:




ANSWER:

x = 32; ;





ANSWER:18 ft 11 in.


8.81 and 4




ANSWER:18

9.25 and 16




ANSWER:20

10.20 and 25




ANSWER:

11.36 and 24




ANSWER:

12.12 and 2.4




ANSWER:

13.18 and 1.5




ANSWER:


14.



ANSWER:

15.



ANSWER:

16.




ANSWER:

17.




ANSWER:

18.

SOLUTION:




ANSWER:

x = 6; ;

19.

SOLUTION:




ANSWER:

; ;

20.

SOLUTION:




ANSWER:

; ;

21.

SOLUTION:




ANSWER:

x 4.7y 1.8z 13.1

22.

SOLUTION:




ANSWER:

x = 40; ;

23.

SOLUTION:




ANSWER:

; ;z = 32





ANSWER:18 ft





ANSWER:161.8 ft


26. and 60




ANSWER:

or 3.5

27. and




ANSWER:

or 0.8

28. and




ANSWER:

or 2.2

Find x, y, and z.

29.

SOLUTION:




ANSWER:

; ; z = 3

30.

SOLUTION:




ANSWER:

; ;





ANSWER:11


SOLUTION:



ANSWER:6.4 ft


SOLUTION:




ANSWER:3.5 ft


SOLUTION:




ANSWER:13.75 ft


35.

SOLUTION:


ANSWER:5

36.

SOLUTION:




ANSWER:8

37.

SOLUTION:




ANSWER:4





SOLUTION:

First find


GF = 65 in.







39.Theorem 8.1

SOLUTION:



.

Prove:

Proof:



of . (Given)



perpendicularlines)


congruent.)


reflexive.)

6. ; AA (Similarity

Statements 4 and 5)


transitive.)

ANSWER:


Prove:








40.Theorem 8.2

SOLUTION:



Prove:









ANSWER:


Prove:





41.Theorem 8.3



Prove: ;

Proof:



(Given)


triangle)





other.)


triangles)

ANSWER:


Prove: ;

Proof:



(Given)


triangle)






triangles)




eSolutions Manual - Powered by Cognero Page 1

8-1 Geometric Mean


1.5 and 20




ANSWER:10

2.36 and 4




ANSWER:12

3.40 and 15




ANSWER:

or 24.5


SOLUTION:



Therefore,

ANSWER:

Find x, y, and z.

5.

SOLUTION:



Solve for z .

ANSWER:

x = 6; ; .

6.

SOLUTION:




ANSWER:

x = 32; ;





ANSWER:18 ft 11 in.


8.81 and 4




ANSWER:18

9.25 and 16




ANSWER:20

10.20 and 25




ANSWER:

11.36 and 24




ANSWER:

12.12 and 2.4




ANSWER:

13.18 and 1.5




ANSWER:


14.



ANSWER:

15.



ANSWER:

16.




ANSWER:

17.




ANSWER:

18.

SOLUTION:




ANSWER:

x = 6; ;

19.

SOLUTION:




ANSWER:

; ;

20.

SOLUTION:




ANSWER:

; ;

21.

SOLUTION:




ANSWER:

x 4.7y 1.8z 13.1

22.

SOLUTION:




ANSWER:

x = 40; ;

23.

SOLUTION:




ANSWER:

; ;z = 32





ANSWER:18 ft





ANSWER:161.8 ft


26. and 60




ANSWER:

or 3.5

27. and




ANSWER:

or 0.8

28. and




ANSWER:

or 2.2

Find x, y, and z.

29.

SOLUTION:




ANSWER:

; ; z = 3

30.

SOLUTION:




ANSWER:

; ;





ANSWER:11


SOLUTION:



ANSWER:6.4 ft


SOLUTION:




ANSWER:3.5 ft


SOLUTION:




ANSWER:13.75 ft


35.

SOLUTION:


ANSWER:5

36.

SOLUTION:




ANSWER:8

37.

SOLUTION:




ANSWER:4





SOLUTION:

First find


GF = 65 in.







39.Theorem 8.1

SOLUTION:



.

Prove:

Proof:



of . (Given)



perpendicularlines)


congruent.)


reflexive.)

6. ; AA (Similarity

Statements 4 and 5)


transitive.)

ANSWER:


Prove:








40.Theorem 8.2

SOLUTION:



Prove:









ANSWER:


Prove:





41.Theorem 8.3



Prove: ;

Proof:



(Given)


triangle)





other.)


triangles)

ANSWER:


Prove: ;

Proof:



(Given)


triangle)






triangles)





1.5 and 20




ANSWER:10

2.36 and 4




ANSWER:12

3.40 and 15




ANSWER:

or 24.5


SOLUTION:



Therefore,

ANSWER:

Find x, y, and z.

5.

SOLUTION:



Solve for z .

ANSWER:

x = 6; ; .

6.

SOLUTION:




ANSWER:

x = 32; ;





ANSWER:18 ft 11 in.


8.81 and 4




ANSWER:18

9.25 and 16




ANSWER:20

10.20 and 25




ANSWER:

11.36 and 24




ANSWER:

12.12 and 2.4




ANSWER:

13.18 and 1.5




ANSWER:


14.



ANSWER:

15.



ANSWER:

16.




ANSWER:

17.




ANSWER:

18.

SOLUTION:




ANSWER:

x = 6; ;

19.

SOLUTION:




ANSWER:

; ;

20.

SOLUTION:




ANSWER:

; ;

21.

SOLUTION:




ANSWER:

x 4.7y 1.8z 13.1

22.

SOLUTION:




ANSWER:

x = 40; ;

23.

SOLUTION:




ANSWER:

; ;z = 32





ANSWER:18 ft





ANSWER:161.8 ft


26. and 60




ANSWER:

or 3.5

27. and




ANSWER:

or 0.8

28. and




ANSWER:

or 2.2

Find x, y, and z.

29.

SOLUTION:




ANSWER:

; ; z = 3

30.

SOLUTION:




ANSWER:

; ;





ANSWER:11


SOLUTION:



ANSWER:6.4 ft


SOLUTION:




ANSWER:3.5 ft

34.Refer t

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eSolutions Manual - Powered by Cognero Page 1 · Find the geometric mean between each pair of numbers. 5 and 20 62/87,21 By the definition, the geometric mean x of any two numbers